Notes21 Protection

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Notes 21: Protection

21.0 Introduction

Distribution system protection involves

using current interrupting devices to de-

energize faulted portions of the network so

that the associated high fault currents will

not flow. Otherwise, continued high faultcurrent that is not interrupted can cause

significant heat, resulting in damaged

equipment (e.g., line burn-downs,

transformer loss of life) and in the worst

case fires and explosions. In addition, faultscan sometimes expose people to live

conductors, resulting in a safety hazard.

In addition, the coordination of protective

devices can be done in such a way so as to

minimize interruptions and interruptiondurations as seem by customer, thus

improving their overall reliability levels.

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In this document, we will briefly survey the

field of protection for distribution systems.

This should be viewed as an introductiononly. Students desiring more in-depth

treatment of the subject should refer to one

of the four very good references listed in the

bibliography of this document.

21.1 Basic principles

Directional relays are unnecessary in radial

distribution systems that do not have any

distributed generation. So almost all

distribution protection devices are simply

non-directional, overcurrent devices.

(Exceptions are, however, when the

distribution system has multiple feeds and/or

when it contains generation).

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The four most basic requirements for an

overcurrent interrupting device is that

1.

Do not operate for highest load current.2. Do operate for minimum fault current.

3. Must be capable of interrupting the

maximum fault current.

4. Interrupt more severe faults more

quickly.

21.1.1 Maximum load current

The load currents of concern are those

highest in magnitude which therefore may

appear to the protective device as a fault

current. An interrupting device should

obviously never open for any load current.

The highest load current is typically the

current that occurs just as the system is re-

energized following an outage. This currentis commonly referred to as the cold-load

pickup current.

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The cold-load pick-up current can be

significantly higher than the normal peak

load current because of two main reasons.

Current in-rush: Cold-load pickup currents

can be affected during the first second or so

following energization by current inrush,

which is the tendency of magnetic loads

(e.g., transformers and motors) to draw highcurrents during the initial transient period as

a result of the difference between the

residual magnetization of the iron and the

initial phase applied across the device.

Simultaneous operation: Cold-load pickup is

also affected by the fact that following an

outage, on re-energizing, many devices that

normally do not run simultaneously will

switch on instantaneously, and

simultaneously. This is particularly true ofdevices that perform temperature control,

e.g., air conditioners, furnaces, water

heaters, refrigerators.

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Cold-load pickup currents can range from 2-5 times normal peak load currents during the

first few seconds following re-energization

and [2].

21.1.2 Minimum fault current

The minimum fault current is determined by

the least severe fault within the protection

zone of the device.

The protection zone is defined by the reach

of the device. The reach is the maximum

distance from the device to a fault for which

the protective device is supposed to operate.

As the fault moves further from the source,

the impedance between source and faultincreases and the fault current decreases.

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So the reach of an overcurrent device

increases as the current interruption setting

decreases.

Observation: Location of the maximum

distance fault depends only on the current

setting and not on the device location.

So basic procedure is that one identifies theleast severe fault type, computes the fault

current for the maximum distance fault. This

fault current is the minimum fault current

and the device should be set to operate at a

level

Significantly higher than the maximum

load current and

Significantly lower than the minimum

fault current

What happens when this is not possible?

Is it ever the case that it is not possible?

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Fact: There always exist faulted conditions

for which the fault current is less than the

maximum load current!!!

How can this be?

Consider the single-line-to-ground (SLG)

fault. In notes20, we found that the

connection for a SLG fault was as in Fig. 1:

0

faI

2

faI

1

faI

1.0

Z1

Z2

Z0

Fig. 1

Sequence fault currents are:

021

021 0.1

ZZZIII fafafa (1)

And the phase a fault current Ifa=3I1

fa.

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However, this is for a so-called bolted fault,

which means that the fault is a pure short-

circuit with no fault impedance.

For a SLG fault through an impedance Zf,

the fault impedance shows up in all three

sequence networks, so that the sequence

network connection is as in Fig. 2.

0

faI

2

faI

1

faI

1.0

Z1

Z2

Z0

3Zf

Fig. 2

Then the sequence fault currents are:

f

fafafa

ZZZZ

III

3

0.1

021

021

(2)

Again we have that Ifa=3I1fa.

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The difference between this case and the

bolted fault case is that the 3Zf term is in the

denominator, so that the fault current isdecreased. If Zf is large, the fault current can

be decreased significantly.

What can Zf be? Consider the case where an

overhead conductor crossing a street breaks

and falls to the pavement. In this case, theimpedance between the conductor and

ground is going to be very high. Then,

clearly, the fault current is going to be very

low.

Such a situation is quite dangerous because

the protection device will not trip.

It is for this reason that a distribution

substation transformer is almost never

connected in ungrounded Y, because then,the zero sequence circuit is open, which is

equivalent to Zf=, and no fault current

flows!

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So we see that, because of the variability of

Zf, there will always be some faults that

result in currents that are below themaximum load current.

Therefore, it is typical to define the

minimum fault current as that which would

flow for a SLG fault through a fault

impedance of 30 or 40 ohms.

21.1.3 Time to operate

The heating effect of a fault current is

depends on the energy dissipated. The

energy dissipated depends on two things:

Power and

Time

Since the power is given by I2R, and since R

is fixed, we see that the variable parameters

for a particular faulted condition are thesquare of the current and the time or

duration of the fault.

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So I2t is an important parameter in

determining the severity of a fault.

In order to protect equipment and people,

protection should be designed so that it

avoids exceeding a certain value of I2t.

So we see that a good design for a protection

device is one that operates faster (smaller t)for higher values of current (larger I).

This thinking results in the so-called time-

inverse characteristic of a protection device,

shown in Fig. 3.

I

t

Fig. 3

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21.2 Types of protection devices

There are four basic types of protectiondevices used in distribution systems. These

are

1. Fuses

2. Automatic circuit reclosers

3. Sectionalizers

4. Circuit breakers and relays

We will discuss each of these briefly.

21.2.1 Fuses

Fuses have elements that melt if enough

current flows through it for enough time.

They are inexpensive, and they can open

very fast for high currents.

The most common type of fuse is the

expulsion type within a cue out, illustrated

in Fig. 4.

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Fig. 4

The purpose of the cutout is to support thefuse and enable efficient replacement when

it is blown. One often sees these devices.

The fuse is called expulsion because, when

the fusible element, located within a fuse

tube, melts under high current, an arcremains. The arc causes a rapid pressure

buildup, which forces the ionized gas out of

the bottom of the cutout, which helps to

prevent reigniting. Fig. 5 shows operation of

an expulsion fuse, which suggests that they

should not be placed where a blast of hot

ionized gas could cause a flashover on

another phase.

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Fig. 5 [2]

Industry standards specify two types of

expulsion fuses.

K type: fast fuse with speed ratio of 6-8

T-type: slower fuse with speed ratio 10-13

The speed ratio is the ratio of

The melting current at 0.1 second to

The melting current at X seconds, where Xis 300 for fuse ratings below 100 amps and

X is 600 for fuse ratings above 100 amps.

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The current rating is the level of current the

fuse can safely carry for an indefinite period

of time.

For example, a typical K-type fuse would

have ratio (using 600 seconds) of

200/30=6.7.

21.2.2 Circuit breakers and relays

21.2.3 Automatic circuit reclosers

ACRs are essentially circuit breakers with

increased capacity both in terms of

interrupting capacity and normal load

carrying capacity.

They are less expensive than circuit breakers

but less accurate.

ACRs are motivated by the fact that 60-80%

of all overhead distribution faults are

temporary and last only a few cycles or a

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few seconds. For example, a branch may

blow against a circuit and then fall to the

ground.

As a result, it becomes very attractive to

reclose following an initial opening of the

protection device.

So the basic operation of an ACR is to Sense the overcurrent

Time and interrupt the overcurrent

Reclose to re-energize the circuit

An ACR can reclose as many as 3-4 times

before locking out.

The first reclose is generally done

instantaneously which means there is no

intentional time delay embedded in the first

reclose. Typical timing sequences used in

the industry include

0-15-30 seconds

0-30-60-90 seconds

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5-45 seconds

But most ACRs can be set for any

combination of instantaneous operations ortime delayed operations.

21.2.4 Sectionalizers

Sectionalizers are switches. The main

difference between a switch an all otherdevices described previously is that all the

other devices can detect and interrupt a

short-circuit fault current, but a switch can

do neither.

It is typically applied together with an ACR

or a circuit breaker to isolate faulted sections

of the distribution system.

So how does a sectionalizer operate?

It counts the number of operations of thebackup device

After a pre-set number of interruptions, it

opens while the backup device is open.

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Then the backup device can reclose to

restore service to remaining circuits.

Advantage relative to fuse is it is

unnecessary to replace it.

Advantage relative to circuit breaker or

ACR is it is less expensive.

Clear disadvantage is that it is a switch.

21.3 Coordination

The basic issue in coordinating multiple

devices in a distribution feeder is that we

want to ensure that only the minimum

number of customers are interrupted for a

specific fault, given the devices that we have

in place.

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We use the terms protected device

(upstream) and protecting device

(downstream).

Important coordinating principles:

1. The protecting device must clear a fault

before the protected device interrupts the

circuit (fuse) or operates to lockout.

2. Outages caused by permanent faultsmust be restricted to the smallest section

of the system for the shortest time.

Figure 6 shows an example of system

coordination where a substation receives

power flow a high voltage transmission line

and steps the voltage down.

Protective devices in this illustration are

located at the coordinating points. Device A

is at the substation. Devices C and H are inthe feeder. Device B is at the branch tap off

the feeder.

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Device D is on the transformer primary and

devices E, F, and G are service entrance

fuses on the transformer secondary. Alldevices must be selected to carry normal

load current and respond properly to a fault.

Fig. 6 [1]

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With respect to device H, device C is the

protected device. For a fault at point 1,

device C must not open and device H mustinterrupt.

With respect to device A, device C is the

protecting device and must interrupt

permanent fault current at point 2 before

device A operates to lockout. Device B isalso a protecting device for A and must

operate similar to device C for a fault at

point 3. Only device A functions when a

fault occurs between A and C, such as at

point 4. For a transformer fault at point 5,

device D is expected to interrupt current and

permit normal load current to flow in the

remainder of the system. For an overload on

the secondary tap of transformer at point 6,

device E should interrupt the circuit,

permitting power to the transformer to becontinued so that customers on other

secondary taps will receive service.

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References

[1] McGraw-Edison Power Systems,Distribution system protection manual.

[2] T. Short, Electric power distribution

handbook, CRC press, 2004.

[3] T. Gonen, Electric Power Distribution

System Engineering, McGraw-Hill, 1986.

[4] J. Blackburn and G. Rockefeller,Applied Protective Relaying,

Westinghouse Corporation.

[5] J. Blackburn, Protective Relaying:

Principles and Applications, 2nd edition,

Marcel-Dekker, New York, 1998.

[6] P. Anderson, Power system protection,

McGraw-Hill, 1999.

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Notes21 Protection